# Talk:Structure set (Ex)

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− | We begin with the map | + | We begin with the map $\mathcal{S}^s(M) \to \mathcal{M}(M)$, from the simple structure set of manifolds simply homotopy equivalent to the orbit space and then we show two things: |

* If manifolds $N$ and $N'$ simply homotopy equivalent to $M$ are diffeomorphic then their images by the map belong to the same orbit of $\mathcal{E}^s(M)$-action on $\mathcal{S}^s(M)$. | * If manifolds $N$ and $N'$ simply homotopy equivalent to $M$ are diffeomorphic then their images by the map belong to the same orbit of $\mathcal{E}^s(M)$-action on $\mathcal{S}^s(M)$. | ||

− | * If two elements of $\mathcal{S}^s(M)$ belong to the same orbit of $\mathcal{E}^s(M)$-action, then | + | * If two elements of $\mathcal{S}^s(M)$ belong to the same orbit of $\mathcal{E}^s(M)$-action, then they are diffeomorphic. |

− | Let $N$ be a smooth manifold and $f\colon N\to M$ a simple homotopy equivalence. Consider a map which takes $N$ to $[(N,f)]\in \mathcal{S}^ | + | Let $N$ be a smooth manifold and $f\colon N\to M$ a simple homotopy equivalence. Consider a map which takes $N$ to $[(N,f)]\in \mathcal{S}^s(M)$. Suppose now that $N'$ is a manifold diffeomorphic to $N$, and $f'\colon N'\to M$ a simple homotopy equivalence (possibly $N'=N$ and $f\nsim f'$). Then there exists $h\colon M\to M$ such that the following diagram commutes. |

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## Revision as of 08:57, 28 August 2013

**Solution**:

We begin with the map , from the simple structure set of manifolds simply homotopy equivalent to the orbit space and then we show two things:

- If manifolds and simply homotopy equivalent to are diffeomorphic then their images by the map belong to the same orbit of -action on .
- If two elements of belong to the same orbit of -action, then they are diffeomorphic.

Let be a smooth manifold and a simple homotopy equivalence. Consider a map which takes to . Suppose now that is a manifold diffeomorphic to , and a simple homotopy equivalence (possibly and ). Then there exists such that the following diagram commutes.

Map is given by composition (the homotopy inverse) and hence is a simple homotopy equivalence. The commutativity of the diagram tells us that in we have the following equalities.

where denotes the -action. Therefore and belong to the same orbit.

Suppose now, that belong to the same orbit of -action. It means, that there exist a simple homotopy equivalence such that But equality in the simple structure set is just the existence of a diffeomorphism making the following diagram commute.